What Is the Acceleration of Masses on a Frictionless Tabletop with Hanging Mass?

[Marioqwe]

Homework Statement

Two masses, m1 = 3.50kg and m2 = 5.00 kg, are on a frictionless tabletop and mass m3 = 7.60kg is hanging from m1. THE COEFFICIENT OF STATIC AND KINETIC FRICTION BETWEEN m1 AND m2 are 0.60 and 0.50 respectively.

a) What are the acceleration of m1 and m2?

Homework Equations

Equation for kinetic and static friction and Newton's second law.

The Attempt at a Solution

I found the acceleration for m1 to be 5.2 m/s^2 by doing

a = ((m3)g - Fk) / (m1 + m3)

Now, I don't know what to do in order to find the acceleration of m2.
Do I have to consider the kinetic friction between m2 and m1 and Newton's third law meaning that m1 exerts a force equal to Fk (force due to kinetic friction) on m2?

 

[cupid.callin]

are m1 and m2 connected by string or what?

 

[Marioqwe]

are m1 and m2 connected by string or what?

No no. m1 is resting on top of m2. And m2 is on top of a frictionless table. I forgot to add the part in capital letters to the problem.

 

[cupid.callin]

you can find the friction acting on m1 due to m2
same will act on m2 and cause acceleration in it

 

[rwisz]

I would approach this problem by first defining the variables and equations involved. In this case, we have three masses (m1, m2, and m3) and two coefficients of friction (static and kinetic). We can also use Newton's second law, F=ma, to calculate the acceleration of each mass.

For m1, we can use the equation you provided, a = ((m3)g - Fk) / (m1 + m3), where g is the acceleration due to gravity (9.8 m/s^2). This equation takes into account the force of gravity acting on m3 and the force of kinetic friction between m1 and m3. However, we also need to consider the force of static friction between m1 and m2.

To find the acceleration of m2, we can use Newton's second law again, F=ma. The forces acting on m2 are the force of static friction from m1 and the force of gravity pulling it down. We can write this as m2a = Fsf - (m2g), where Fsf is the force of static friction and m2g is the force of gravity. We can also use the definition of static friction, Fsf = μsN, where μs is the coefficient of static friction and N is the normal force between m1 and m2. We can then substitute this into our equation to get m2a = μsN - (m2g).

To find the normal force, we can use the fact that the table is frictionless, so the net force in the vertical direction must be zero. This means that N = m2g. Substituting this into our equation, we get m2a = μsm2g - (m2g), which simplifies to m2a = (μs - 1)m2g. Finally, we can solve for a by dividing both sides by m2, giving us a = (μs - 1)g.

Therefore, the acceleration of m2 is (μs - 1)g = (0.60 - 1)(9.8 m/s^2) = -4.9 m/s^2. This negative sign indicates that the acceleration of m2 is in the opposite direction of the force of gravity, which makes sense since the force of static friction is acting in the opposite direction to prevent m2 from moving.